Simplify and expand the following expression: $ \dfrac{2y - 3}{2y - 5}-\dfrac{y - 6}{4y + 10} $
Explanation: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(2y - 5)(4y + 10)$ Multiply the first term by $\dfrac{4y + 10}{4y + 10}$ $ \begin{align*} \dfrac{2y - 3}{2y - 5} \times \dfrac{4y + 10}{4y + 10} & = \dfrac{(2y - 3)(4y + 10)}{(2y - 5)(4y + 10)} \\ & = \dfrac{8y^2 + 8y - 30}{(2y - 5)(4y + 10)}\end{align*} $ Multiply the second term by $\dfrac{2y - 5}{2y - 5}$ $ \begin{align*} \dfrac{y - 6}{4y + 10} \times \dfrac{2y - 5}{2y - 5} & = \dfrac{(y - 6)(2y - 5)}{(4y + 10)(2y - 5)} \\ & = \dfrac{2y^2 - 17y + 30}{(4y + 10)(2y - 5)}\end{align*} $ Now we have: $ = \dfrac{8y^2 + 8y - 30}{(2y - 5)(4y + 10)} - \dfrac{2y^2 - 17y + 30}{(4y + 10)(2y - 5)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{8y^2 + 8y - 30 - (2y^2 - 17y + 30)}{(2y - 5)(4y + 10)} $ $ = \dfrac{8y^2 + 8y - 30 - 2y^2 + 17y - 30}{(2y - 5)(4y + 10)} $ $ = \dfrac{6y^2 + 25y - 60}{(2y - 5)(4y + 10)}$ Expand the denominator: $ = \dfrac{6y^2 + 25y - 60}{8y^2 - 50}$